$h(x) = 5x^{3}+4x^{2}-3-2(f(x))$ $g(t) = 3t^{3}-5t^{2}+5t-4(h(t))$ $f(n) = 5n^{3}+6n^{2}$ $ g(f(0)) = {?} $
Solution: First, let's solve for the value of the inner function, $f(0)$ . Then we'll know what to plug into the outer function. $f(0) = 5(0^{3})+6(0^{2})$ $f(0) = 0$ Now we know that $f(0) = 0$ . Let's solve for $g(f(0))$ , which is $g(0)$ $g(0) = 3(0^{3})-5(0^{2})+(5)(0)-4(h(0))$ To solve for the value of $g$ , we need to solve for the value of $h(0)$ $h(0) = 5(0^{3})+4(0^{2})-3-2(f(0))$ To solve for the value of $h$ , we need to solve for the value of $f(0)$ $f(0) = 5(0^{3})+6(0^{2})$ $f(0) = 0$ That means $h(0) = 5(0^{3})+4(0^{2})-3+(-2)(0)$ $h(0) = -3$ That means $g(0) = 3(0^{3})-5(0^{2})+(5)(0)+(-4)(-3)$ $g(0) = 12$